If P=(10121) then P2022=

P=(10121) ⇒P2=(10121) (10121)=(1011) P3=(1011)(10121)=(10321)P4=(10321) (10121)=(1021)Similarly P2022=(1010111)

If P=(10121) then P2022=

P=(10121) ⇒P2=(10121) (10121)=(1011) P3=(1011)(10121)=(10321)P4=(10321) (10121)=(1021)Similarly P2022=(1010111)

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If $P = (\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix})$ then $P^{2022} =$

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$(\begin{matrix} 1 & 0 \\ 2022 & 1 \end{matrix})$

$(\begin{matrix} 1 & 0 \\ 2011 & 1 \end{matrix})$

$(\begin{matrix} 1 & 0 \\ 1011 & 1 \end{matrix})$

$(\begin{matrix} 1 & 0 \\ 1000 & 1 \end{matrix})$

answer is C.

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Detailed Solution

$P = (\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}) \Rightarrow P^{2} = (\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})$

$P^{3} = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}) = (\begin{matrix} 1 & 0 \\ \frac{3}{2} & 1 \end{matrix})$

$P^{4} = (\begin{matrix} 1 & 0 \\ \frac{3}{2} & 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ \frac{1}{2} & 1 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix})$